Question

1375. Number of Times Binary String Is Prefix-Aligned

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Description

You have a 1-indexed binary string of length n where all the bits are 0 initially. We will flip all the bits of this binary string (i.e., change them from 0 to 1) one by one. You are given a 1-indexed integer array flips where flips[i] indicates that the bit at index i will be flipped in the ith step.

A binary string is prefix-aligned if, after the ith step, all the bits in the inclusive range [1, i] are ones and all the other bits are zeros.

Return _the number of times the binary string is prefix-aligned during the flipping process_.

 

Example 1:

Input: flips = [3,2,4,1,5]
Output: 2
Explanation: The binary string is initially "00000".
After applying step 1: The string becomes "00100", which is not prefix-aligned.
After applying step 2: The string becomes "01100", which is not prefix-aligned.
After applying step 3: The string becomes "01110", which is not prefix-aligned.
After applying step 4: The string becomes "11110", which is prefix-aligned.
After applying step 5: The string becomes "11111", which is prefix-aligned.
We can see that the string was prefix-aligned 2 times, so we return 2.

Example 2:

Input: flips = [4,1,2,3]
Output: 1
Explanation: The binary string is initially "0000".
After applying step 1: The string becomes "0001", which is not prefix-aligned.
After applying step 2: The string becomes "1001", which is not prefix-aligned.
After applying step 3: The string becomes "1101", which is not prefix-aligned.
After applying step 4: The string becomes "1111", which is prefix-aligned.
We can see that the string was prefix-aligned 1 time, so we return 1.

 

Constraints:

• n == flips.length
• 1 <= n <= 5 * 104
• flips is a permutation of the integers in the range [1, n].

Solutions

Solution 1: Direct Traversal

We can traverse the array $flips$, keeping track of the maximum value $mx$ of the elements we have traversed so far. If $mx$ equals the current index $i$ we are traversing, it means that the first $i$ elements have all been flipped, i.e., the prefix is consistent, and we increment the answer.

After the traversal is finished, we return the answer.

The time complexity is $O(n)$, where $n$ is the length of the array $flips$. The space complexity is $O(1)$.

class Solution:
  def numTimesAllBlue(self, flips: List[int]) -> int:
    ans = mx = 0
    for i, x in enumerate(flips, 1):
      mx = max(mx, x)
      ans += mx == i
    return ans

class Solution {
  public int numTimesAllBlue(int[] flips) {
    int ans = 0, mx = 0;
    for (int i = 1; i <= flips.length; ++i) {
      mx = Math.max(mx, flips[i - 1]);
      if (mx == i) {
        ++ans;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int numTimesAllBlue(vector<int>& flips) {
    int ans = 0, mx = 0;
    for (int i = 1; i <= flips.size(); ++i) {
      mx = max(mx, flips[i - 1]);
      ans += mx == i;
    }
    return ans;
  }
};

func numTimesAllBlue(flips []int) (ans int) {
  mx := 0
  for i, x := range flips {
    mx = max(mx, x)
    if mx == i+1 {
      ans++
    }
  }
  return
}

function numTimesAllBlue(flips: number[]): number {
  let ans = 0;
  let mx = 0;
  for (let i = 1; i <= flips.length; ++i) {
    mx = Math.max(mx, flips[i - 1]);
    if (mx === i) {
      ++ans;
    }
  }
  return ans;
}